Question

我需要帮助来解决这个问题：我必须计算更深的分支的长度，该分支仅包含元素＆＃34; A＆＃34;在BSTree。

我试过这个：

data El = A | B deriving (Eq,Show)

data BSTree a = Nil | Node a (BSTree a) (BSTree a)
testBST :: BSTree El
testBST = Node A (Node A (Node B Nil Nil) Nil) (Node A Nil (Node A (Node B Nil Nil) Nil)) 

pathTree_hlp :: Int -> Int -> BSTree El -> Int
pathTree_hlp curmax prevmax Nil = max curmax prevmax
pathTree_hlp curmax prevmax (Node A l r) = pathTree_hlp (curmax+1) prevmax l + pathTree_hlp (curmax+1) prevmax r 
pathTree_hlp curmax prevmax (Node B l r) = pathTree_hlp 0 (max curmax prevmax) l + pathTree_hlp 0 (max curmax prevmax) r

pathInTree :: BSTree El -> Int
pathInTree Nil = 0
pathInTree (Node B l r) = 0
pathInTree (Node A l r) = pathTree_hlp 0 0 (Node A l r)

结果应该是3，但我的代码返回17（？）你能帮我解决这个问题吗？

感谢所有人。

Answer 1

在这两行中，我改变了函数的 sum ，其中l和r分支为 max ：

pathTree_hlp curmax prevmax (Node A l r) = max (pathTree_hlp (curmax+1) prevmax l) (pathTree_hlp (curmax+1) prevmax r)
pathTree_hlp curmax prevmax (Node B l r) = max (pathTree_hlp 0 (max curmax prevmax) l) (pathTree_hlp 0 (max curmax prevmax) r)

计算Tree中更深的分支

1 个答案: